Modeling Probabilities In Games Of Tennis Biology Essay

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This internal assessment will investigate probabilities in tennis games. These probabilities will be dependent on the chances of one of the two players scoring one point in a game. Based on that, probability models will be developed

Considering the two players, Adam and Ben, Adam has the probability of scoring twice as much points as Ben would. Based on this we could begin the modeling. Let the random variable p represent the probability of Adam winning. Based on the information we are given:

p =P(A)=

Assuming that Ben winning would be q, which means Adam does not win we can calculate the chances of Ben winning as follows:

Due to the probability having two values only, Adam will either win or lose. Based on this, we will use the binomial distribution method to form the probability distribution table. The binomial equation is as follows:

P(X) = nCX pX qn-X

P(X) is the probability of any score. n is the maximum number of points, which in this case is 10. As mentioned above, p is , which is the probability of Adam winning, while q is , which is the probability of Ben winning. X takes one of the following values 0,1,2,3… 10. Substituting these values into the binomial equation will give the following:

P(X) = 10CX X 10-X

Before tabulating the results, one has to wonder their validity. The data that was given says that the players have been against each other often enough. Mentioning often enough does not really make the data valid. In order for develop a completely valid mathematical model of this, the players should play against each other infinity number of times. Therefore, we could assume "often enough" is near infinity. Also something to question the validity is the fact that they are human. One player might be better than the other in different conditions, whether these conditions are physical or mental. These are certainly factors that should be seen when considering validity.

The tabulated results of the probabilities:

Xi

P(Xi)

0

 10C0010-0

= 0.000017

1

 10C1110-1

= 0.000339

2

 10C2210-2

= 0.003048

3

 10C3310-3

= 0.016258

4

 10C4410-4

= 0.056902

5

 10C5510-5

= 0.136565

6

10C6610-6

= 0.227608

7

10C7710-7

= 0.260123

8

10C8810-8

 = 0.195092

9

10C9910-9

= 0.086708

10

10C101010-10

= 0.017342 

To make sure these probabilities are correct, adding them up should result to 1.

Graphically representing the tabulated data in a histogram, using Microsoft Excel:

Expected Value= Mean= E(X) = np

Standard Deviation = =

E(X) = 10 x =

= = = 1.4907

The expected value, which is calculated to be 6.67, gives us an indication that Adam has a chance of winning 6 or 7 points out of 10. The value of the standard deviation states to us that Adam may score 8 points or 5 also as the standard deviation shows the deviation away from the mean upwards or downwards by 1.4907.

Part II: Non-extended play games

Assuming there is a game between Adam and Ben under official tennis rules. Based on what was calculated and provided from data, the probability of Adam winning a point during such games is , while Ben's probability of winning a point is . Using this data, we will develop a model on the chances of Adam and Ben winning this best out of seven point game. Assuming that Y is the sum of points scored during the game, the following possibilities for the values of Y can be shown:

4,5,6,7

4-0, 4-1, 4-2, 4-3

These are the expected results in case of Adam or Ben winning. For the possibility of Adam winning 4-0, there is only one possible combination of points during the game which can be shown as follows:

Adam scores, Adam scores, Adam scores, Adam scores and Adam scores

We will denote this as AAAA, for that A depicts a point scored by Adam while B depicts a point scored by Ben. So assuming that Ben won 4-0, we would say BBBB. This shows so far only two possibilities for the outcome of the games.

With the possibility of 4-0 now aside, I will discuss the other possibilities. Since the winner of the game has to score the final point of the game, even if it goes into deuce, then the three points he scored before can be in any combination. For example, Adam would win points up to 3-0, but then Ben scores twice for the score to become 3-2. Adam eventually would win the game 4-2. In this game we would say that it went like this: AAABBA. No matter what is the case, Adam has to score the final point if he is to be the winner. The previous combination can look like this: BAAABA. Still Adam wins, but the combination of the first three points was different. Based on this, we would develop the following model of combination:

Y-1C3

The Y-1 part of the model is the numerical order of the last point scored in the game. Therefore, based on this model we can calculate the combinations of ways to win for Adam can be calculated:

AAAA (i.e. 4C4) + 5-1C3 + 6-1C3+ 7-1C3=

1 + 4 + 10 + 20 = 35

The combination of ways to win for Ben:

BBBB (i.e. 4C4) + 5-1C3 + 6-1C3+ 7-1C3= 35

This means that there are 70 possible ways for this game to end.

Combining the model that was just developed with the binomial probability theorem, we can calculate the probability of Adam winning a game based on the equation mentioned above:

P(X) = nCX pX qn-X

P(X) = YC4 p4 qY-4 (this is just in case Adam wins 4-0)

P(X) = Y-1C3 p4 qY-4

These two equations the probability of Adam winning can be shown as follows:

Score

Probability

4-0

 4C4(4(0

=

4-1 

 5-1C3(4(1

=

4-2

6-1C3(4(2

 =

4-3 

7-1C3(4(3

 =

Probability of Adam winning= + + + =

Logically, the probability of Ben winning is 1 minus the probability of Adam winning since the sum of all the probabilities adds to one. Therefore:

Probability of Ben winning= 1 - =

Based on all what was done above, we can easily generalize a model that would fit into this investigation. We can represent the sum of different probabilities for the equation Y-1C3 p4 qY-4 by adding a sigma sign before it. Therefore, the general model would like:

Y-1C3 p4 qY-4 + p4 q0

This can be represented in the form of c (winning) and d (losing) as follows:

Y-1C3 c4 dY-4 + c4 d0

Since d0 is equal to one, a more simplified version of the model is as follows:

Y-1C3 c4 dY-4 + c4

Part III: Extended play games

With the games being endless, there is a chance that the game can go into infinity. This means that the values of Y can be greater than or equal to four. Still, assuming the game does not end with a deuce, the values of Y will be from 4 to 6. Therefore, the non-deuce games can be calculated as follows:

Probability of Adam winning 4-0:

P(A) = YC4 p4 qY-4 = 4C4 (4 (4-4 =

Probability of Adam winning a game in 5 points:

P(A) = Y-1C3 p4 qY-4 = 5-1C3 (4 (1 =

Probability of Adam winning a game in 6 points:

P(A) = Y-1C3 p4 qY-4 = 6-1C3 (4 (2 =

Probability of Adam winning non-deuce games:

P(A) = + + =

Probability of Ben winning 4-0:

P(B) = YC4 p4 qY-4 = 4C4 (4 (4-4 =

Probability of Ben winning a game in 5 points:

P(B) = Y-1C3 p4 qY-4 = 5-1C3 (4 (1 =

Probability of Ben winning a game in 6 points:

P(B) = Y-1C3 p4 qY-4 = 6-1C3 (4 (2 =

Probability of Ben winning non-deuce games:

P(B) = + + =

We cannot calculate a game that ends at Y=7 due to the fact that it would be a deuce and deuce games end only with a two point advantage, which means that beyond Y=6 the game only ends when Y is an even number. This creates a probability that the game could be infinite.

Now assuming that a deuce would be called, the first deuce requires all combinations of Y=6 other than 4-2. For the deuce to be called on, each player has to score a point (AB or BA). The win requires two consecutive points (AA or BB). The following shows an assumption of how a game could be infinite:

AAABBB((n-1)(AB)or (n-1)(BA))

In this case, it is all the combinations of AAABBB. n, in this case, is the deuce number being called after the first deuce. Meaning all values of n are greater than 1. n>1. Based on this, we could see that the probability of a deuce occurring takes a form of a geometric series. With this, we could find the sum to infinity of the probabilities. The formula of the sum to infinity of a geometric series is as follows.

The probability of the first deuce happening (u1):

6C3 (3 (3 =

Probability of Adam winning this deuce:

P(A)= Deuce x (AA) = x x =

Probability of Ben winning this deuce:

P(B)= Deuce x (BB) = x x =

After finding the probability of the first deuce occurring, we should aim to find the probability of others occurring since the conditions required for other deuces are different from the first deuce. The after the first deuce, the number of points are to be subtracted by 8, which represents the number of points after the first deuce. Therefore any other points scored after the first deuce are Y-8. Now, we can combine the probability of deuces to calculate the probability of any Adam winning games with deuces is as follows:

6C3 (3 (3(Y-8 (Y-8(Advantage points)Y-8(pp)  

6C3 (3 (3(Y-8 (Y-8(2)Y-8(2 

For the probability of Adam winning the game itself, we should add the probability of him winning non-deuce games. Therefore:

P(A)= 6C3 (3 (3(Y-8 (Y-8(2)Y-8(2 +

Based on this:

P(B)= 6C3 (3 (3(Y-8 (Y-8(2)Y-8(2 +

Based on the equations developed above we can get a rough probability of both men winning the tennis game:

Y

P(A) - (Non-deuce game)

P(B) - (Non-deuce game)

8

0.097546106

0.0243865264

9

0.043353825

0.0108384562

10

0.019268367

0.0048170916

11

0.008563718

0.0021409296

12

0.003806097

9.515242752E-04

13

0.001691599

4.228996779E-04

14

0.000751822

1.879554124E-04

15

0.000334143

8.353573884E-05

16

0.000148508

3.712699504E-05

17

6.600354674E-05

1.650088668E-05

18

2.933490966E-05

7.333727415E-06

19

1.303773763E-05

3.259434407E-06

20

5.794550056E-06

1.448637514E-06

22

1.144602480E-06

2.861506201E-07

24

2.260943171E-07

5.652357927E-08

26

4.466060585E-08

1.116515146E-08

28

8.821848068E-09

2.205462017E-09

30

1.742587273E-09

4.356468180E-10

34

6.799304100E-11

1.699826020E-11

38

2.652982550E-12

6.632456370E-13

42

1.035152460E-13

2.587881160E-14

50

1.575956320E-16

3.939890790E-17

58

2.399297110E-19

5.998242770E-20

60

4.739352310E-20

1.184838080E-20

Total

0.175579781

0.0438949451

P(A) 0.175579781 + = 0.8559638688

P(B) 0.0438949451 + = 0.1440321193

P(A) : P(B)

0.8559638688 : 0.1440321193

5.942867972 : 1

This shows that the odds of Adam winning are about 6:1 to the of Ben.

Now assuming again Players C and D. Similarly to what we mentioned above about them, we can derive a formula for the probability of one of C winning an extended play game:

Y-1C3 c4 dY-4+ c4 + 6C3 (3 (3(Y-8 (Y-8(2)Y-8(

With this model, we can write out the following spreadsheet based on c= 0.5, 0.55, 0.6, 0.7, 0.9. P(C)= Total + Y-1C3 c4 dY-4:

Y

c= 0.5

c= 0.55

c= 0.6

c= 0.7

c= 0.9

8

0.078125

0.091723577

0.0995328

0.0907578

0.0118098

9

0.0390625

0.045403171

0.047775744

0.038118276

0.002125764

10

0.01953125

0.02247457

0.022932357

0.016009676

0.000382638

11

0.009765625

0.011124912

0.011007531

0.006724064

6.8875E-05

12

0.004882813

0.005506831

0.005283615

0.002824107

1.2398E-05

13

0.002441406

0.002725882

0.002536135

0.001186125

2.2315E-06

14

0.001220703

0.001349311

0.001217345

4.9817E-04

4.0168E-07

15

6.1035E-04

6.6791E-04

5.8433E-04

2.0923E-04

7.2302E-08

16

3.0518E-04

3.3062E-04

2.8048E-04

8.7878E-05

1.3014E-08

17

1.5259E-04

1.6365E-04

1.3463E-04

3.6908E-05

2.3426E-09

18

7.6294E-05

8.1009E-05

6.4622E-05

1.5502E-05

4.2167E-10

20

1.9074E-05

1.9849E-05

1.4889E-05

2.7345E-06

1.3662E-11

22

4.7684E-06

4.8636E-06

3.4304E-06

4.8236E-07

4.4265E-13

24

1.1921E-06

1.1917E-06

7.9036E-07

8.5089E-08

1.4342E-14

26

2.9802E-07

2.9199E-07

1.8210E-07

1.5010E-08

4.6467E-16

28

7.4506E-08

7.1546E-08

4.1956E-08

2.6477E-09

1.5055E-17

30

1.8627E-08

1.7531E-08

9.6666E-09

4.6706E-10

4.8780E-19

34

1.1642E-09

1.0525E-09

5.1314E-10

1.4533E-11

5.1207E-22

38

7.2760E-11

6.3188E-11

2.7240E-11

4.5223E-13

5.3755E-25

42

4.5475E-12

3.7937E-12

1.4460E-12

1.4072E-14

5.6430E-28

50

1.7764E-14

1.3674E-14

4.0747E-15

1.3626E-17

6.2185E-34

60

1.7347E-17

1.2077E-17

2.6455E-18

2.3272E-21

2.2203E-41

Total

0.156199138

0.181577729

0.191368933

0.156471056

0.014402196

Y-1C3c4dY-4

0.339019775

0.436280288

0.539567083

0.742291688

0.984147278

P(C):

0.495218913

0.617858017

0.730936016

0.898762744

0.998549474

In order to calculate the odds, we divide the probability of something happening by the probability of it not happening. In this case we're going to calculate for each case of c.

The odds when c=0.5: = 0.9810567903

The odds when c=0.55: = 1.616828416

The odds when c=0.6: = 2.716588096

The odds when c=0.7: = 8.877786494

The odds when c=0.9: = 688.3779949

Based on the data calculated above, I used a program called Graph to plot a graph:

From the data and the graph above, we can see that there is an exponential relationship between odds and probabilities. This is due to the fact that when the players with similar probabilities (c=0.5,d=0.5) they have the same odds of winning (1:1). As the probability of c increases from 0.5 to 0.55, we see the odds slightly increase but the increase from 0.55 to 0.7 was about 6:1. When the probability went from 0.7 to 0.9 we notice the drastic exponential relation as the odds went from approximately 6 against 1 to 600 against 1, 100 times more than it was at 0.7. This is clear evidence that the relationship is exponential. To support this, the graph shows an exponential relation. The line in the graph represents an exponential line of best fit. We can see that it says R2. R2 represents the correlation regression squared. The correlation regression shows how close the data plotted is to the graph. The correlation regression is , which is 0.9347726996. With a correlation regression of about 0.9, we see that the points plotted are heavily in correlation. Therefore, we can clearly see the exponential relation.

Limitations:

As I have mentioned before, the validity of the models developed is to be debated. The fact that the game can go on for very long makes me point something. If Adam is to be said to have the better talent while Ben has a higher stamina, then clearly Ben will have the greater probability at some point when Y becomes really large and the amount of deuces increase. Another limitation is the probability that weather change could occur during the games and one player plays better than the other in certain weathers. To improve such limitations if further, deeper investigation is to be done, using the Poisson theory would be useful and accurate as it takes into consideration such factors.

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